Intersection of Relations

For example, let \(R\) and \(S\) be the relations “is a friend of” and “is a work colleague of” defined on a set of people \(A\) (assuming \(A = B\)). Their intersection \(R \cap S\) will be the relation “is a friend and work colleague of“.

If the relations \(R\) and \(S\) are defined by matrices \({M_R} = \left[ {{a_{ij}}} \right]\) and \({M_S} = \left[ {{b_{ij}}} \right],\) the matrix of their intersection \(R \cap S\) is given by

Union of Relations

For example, the union of the relations “is less than” and “is equal to” on the set of integers will be the relation “is less than or equal to“.

If the relations \(R\) and \(S\) are defined by matrices \({M_R} = \left[ {{a_{ij}}} \right]\) and \({M_S} = \left[ {{b_{ij}}} \right],\) the union of the relations \(R \cup S\) is given by the following matrix:

Sometimes the converse relation is also called the inverse relation and denoted by \(R^{-1}.\)

Empty, Universal and Identity Relations

A relation \(R\) between sets \(A\) and \(B\) is called an empty relation if \(\require{AMSsymbols}{R = \varnothing.}\)

The universal relation between sets \(A\) and \(B,\) denoted by \(U,\) is the Cartesian product of the sets: \(U = A \times B.\)

A relation \(R\) defined on a set \(A\) is called the identity relation (denoted by \(I\)) if \(I = \left\{ {\left( {a,a} \right) \mid \forall a \in A} \right\}.\)

Properties of Combined Relations

When we apply the algebra operations considered above we get a combined relation. The original relations may have certain properties such as reflexivity, symmetry, or transitivity. The question is whether these properties will persist in the combined relation? The table below shows which binary properties hold in each of the basic operations.

To find the intersection \(R \cap S,\) we multiply the corresponding elements of the matrices \(M_R\) and \(M_S\). This operation is called Hadamard product and it is different from the regular matrix multiplication. So, we have

Example 4.

Let \(B = \left\{ {a,b,c,d} \right\}.\) The relation \(S\) on set \(B\) is defined by the digraph.
Find the combined relation \(\overline {S \cap {S^T}},\) where \({S^T}\) denotes the converse relation of \(S.\)

Solution.

The converse relation \(S^T\) is represented by the digraph with reversed edge directions.

Find the intersection of \(S\) and \(S^T:\)

The complementary relation \(\overline {S \cap {S^T}} \) has the form

Example 5.

Prove that the symmetric difference of two reflexive relations is irreflexive.

Solution.

Let \(R\) and \(S\) be relations defined on a set \(A.\)

Since \(R\) and \(S\) are reflexive we know that for all \(a \in A,\) \(\left( {a,a} \right) \in R\) and \(\left( {a,a} \right) \in S.\)

The difference of the relations \(R \backslash S\) consists of the elements that belong to \(R\) but do not belong to \(S\). Hence, \(R \backslash S\) does not contain the diagonal elements \(\left( {a,a} \right),\) i.e. it is irreflexive.

Similarly, we conclude that the difference of relations \(S \backslash R\) is also irreflexive.

By definition, the symmetric difference of \(R\) and \(S\) is given by

So we need to prove that the union of two irreflexive relations is irreflexive. Suppose that this statement is false. If the union of two relations is not irreflexive, its matrix must have at least one \(1\) on the main diagonal. This is only possible if either matrix of \(R \backslash S\) or matrix of \(S \backslash R\) (or both of them) have \(1\) on the main diagonal. However this contradicts to the fact that both differences of relations are irreflexive. Thus the proof is complete.

We conclude that the symmetric difference of two reflexive relations is irreflexive.

Example 6.

Prove that the union of two antisymmetric relations need not be antisymmetric.

Solution.

We can prove this by means of a counterexample.

Consider the set \(A = \left\{ {0,1} \right\}\) and two antisymmetric relations on it: